Introduction to Linear Algebra: Defining Linear Transformations

Linearity is the skeletal structure of vector spaces. A linear transformation is not just a function; it is a mapping $T$ between vector spaces that honors the fundamental operations of vector addition and scalar multiplication. Think of it as a "structural blueprint"—if you know how the transformation affects a basic set of vectors, you know how it affects the entire universe of those vectors.

The Two Pillars of Linearity

For a transformation $T$ to be considered linear, it must satisfy two strict algebraic conditions for all vectors $v, w$ and all scalars $c$:

Additivity: $T(v + w) = T(v) + T(w)$. The transformation of a sum is the sum of the transformations.
Homogeneity: $T(cv) = cT(v)$. Scaling the input scales the output by the exact same factor.

The Superposition Principle

Combining these rules gives us the most powerful identity in linear algebra:

$$T(c_1v_1 + \dots + c_nv_n) = c_1T(v_1) + \dots + c_nT(v_n)$$

This means that a linear transformation $T$ acts on a linear combination of vectors by distributing across the sum and pulling out the scalars.

The Zero Vector Constraint

A critical "litmus test" for linearity is the Origin Test. If a transformation is linear, it must map the zero vector to the zero vector:

$T(\mathbf{0}) = \mathbf{0}$

If a mapping shifts the origin (e.g., $T(v) = v + b$), it is an affine transformation, not a linear one. In the geometry of the plane, linear transformations keep the center fixed; they never "slide" the space.

Recognizing Non-Linearity

Linearity is incredibly fragile. If the rule governing $T$ involves any of the following, it is not linear:

Squares or higher powers (e.g., $v_1^2$)
Products of components (e.g., $v_1 v_2$)
Absolute values or norms (e.g., $||v||$)
Constant offsets (e.g., $v_1 + 1$)

🎯 Core Principle: Example Contrast

Consider a fixed vector $a = (1, 3, 4)$. The dot product $T(v) = a \cdot v$ is linear because it distributes over addition. However, the norm $T(v) = ||v||$ is not linear; it fails the triangle inequality ($||v+w|| \leq ||v||+||w||$ is not equality) and fails for negative scalars ($||-v|| = ||v|| \neq -||v||$).

QUESTION 1

Which of these transformations from $\mathbb{R}^2$ to $\mathbb{R}$ are NOT linear?

$T(v) = (v_2, v_1)$

$T(v) = v_1 v_2$

$T(v) = v_1 - v_2$

$T(v) = (0, v_1)$

QUESTION 2

Suppose $T$ is reflection across the $x$ axis and $S$ is reflection across the $y$ axis in the $xy$ plane. If $v = (x, y)$, what is the result of the product transformation $S(T(v))$?

$(x, y)$ (The Identity)

$(-x, y)$ (Y-reflection)

$(-x, -y)$ (Reflection through origin)

$(y, x)$ (Reflection across $y=x$)

QUESTION 3

If $u = [1, 0]^T$ and $v = [0, 1]^T$, and we know $Au$ and $Av$ are the columns of matrix $A$. For $w = [5, 7]^T$, how is $Aw$ connected to the columns of $A$?

$Aw = 5Au + 7Av$

$Aw = Au + Av$

$Aw = 7Au + 5Av$

$Aw = [5, 7]^T$

QUESTION 4

The transformation $T(M) = AM$ where $A$ is a fixed $2 \times 2$ matrix and $M$ is any $2 \times 2$ matrix is linear. Which matrix properties prove this?

Determinant rules $|AM| = |A||M|$

Distributive property $A(B+C) = AB+AC$ and scalar associativity $A(cM) = c(AM)$

Commutativity $AM = MA$

The fact that $A$ is invertible

QUESTION 5

If a vector space consists of polynomials of degree 3 or less, why is the matrix $B^2$ (representing the fourth derivative) equal to the zero matrix?

Because the derivative of zero is zero

Because the fourth derivative of any polynomial of degree $\leq 3$ is zero

Because the basis vectors are orthogonal

Because $B$ is a diagonal matrix